Question: Divide the following complex numbers. $ \dfrac{-8-2i}{5-3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+3i}$ $ \dfrac{-8-2i}{5-3i} = \dfrac{-8-2i}{5-3i} \cdot \dfrac{{5+3i}}{{5+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-8-2i) \cdot (5+3i)} {(5-3i) \cdot (5+3i)} = \dfrac{(-8-2i) \cdot (5+3i)} {5^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-8-2i) \cdot (5+3i)} {(5)^2 - (-3i)^2} = $ $ \dfrac{(-8-2i) \cdot (5+3i)} {25 + 9} = $ $ \dfrac{(-8-2i) \cdot (5+3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-8-2i}) \cdot ({5+3i})} {34} = $ $ \dfrac{{-8} \cdot {5} + {-2} \cdot {5 i} + {-8} \cdot {3 i} + {-2} \cdot {3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-40 - 10i - 24i - 6 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-40 - 10i - 24i + 6} {34} = \dfrac{-34 - 34i} {34} = -1-i $